The transverse axis of the hyperbola `5x^2-4y^2-30x-8y+121=0` is

Options

A. `y+1=0`
B. `x-3=0`
C. `x+3=0`
D. `y-1=0`

Answer

To find the transverse axis of the hyperbola given by the equation \(5x^2 - 4y^2 - 30x - 8y + 121 = 0\), we will first convert this equation into its standard form. Here’s a step-by-step solution:

### Step 1: Rearranging the equation
Start with the given equation:
\[ 5x^2 - 4y^2 - 30x - 8y + 121 = 0 \]
We can rearrange it to isolate the constant on one side:
\[ 5x^2 - 30x - 4y^2 - 8y = -121 \]

### Step 2: Completing the square for \(x\)
For the \(x\) terms \(5x^2 - 30x\), we can factor out the 5:
\[ 5(x^2 - 6x) \]
Now, we complete the square inside the parentheses:
\[ x^2 - 6x = (x - 3)^2 - 9 \]
Thus,
\[ 5((x - 3)^2 - 9) = 5(x - 3)^2 - 45 \]

### Step 3: Completing the square for \(y\)
For the \(y\) terms \(-4y^2 - 8y\), we factor out \(-4\):
\[ -4(y^2 + 2y) \]
Now, we complete the square:
\[ y^2 + 2y = (y + 1)^2 - 1 \]
Thus,
\[ -4((y + 1)^2 - 1) = -4(y + 1)^2 + 4 \]

### Step 4: Substitute back into the equation
Now substitute back into the rearranged equation:
\[ 5(x - 3)^2 - 45 - 4(y + 1)^2 + 4 = -121 \]
Combine the constants:
\[ 5(x - 3)^2 - 4(y + 1)^2 - 41 = -121 \]
Add 41 to both sides:
\[ 5(x - 3)^2 - 4(y + 1)^2 = -80 \]

### Step 5: Divide by -80
To convert to standard form, divide the entire equation by -80:
\[ -\frac{5(x - 3)^2}{80} + \frac{4(y + 1)^2}{80} = 1 \]
This simplifies to:
\[ \frac{(y + 1)^2}{20} - \frac{(x - 3)^2}{16} = 1 \]

### Step 6: Identify the standard form
The standard form of the hyperbola is:
\[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \]
where \((h, k)\) is the center, \(a^2 = 20\), and \(b^2 = 16\). Here, \(h = 3\) and \(k = -1\).

### Step 7: Determine the transverse axis
For hyperbolas of the form \(\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1\), the transverse axis is vertical. The equation of the transverse axis is given by:
\[ y = k \]
Thus, for our hyperbola:
\[ y = -1 \]

### Final Answer
The transverse axis of the hyperbola is:
\[ y + 1 = 0 \]

---